Question

Consider 1.40 mol of carbon monoxide and 3.20 mol of chlorine sealed in a 8.00 L container at 476 oC. The equilibrium constant, Kc, is 2.50 (in M-1) for

CO(g) + Cl2(g) ↔ COCl2(g)

Calculate the equilibrium molar concentration of CO.

So I tried this question by finding the equilibrium concentrations then using the quadratic equation to get .187 which wasn't right. Help?


Chemists studied the formation of phosgene by sealing 0.74 atm of carbon monoxide and 0.92 atm of chlorine in a reactor at a certain temperature. The pressure dropped smoothly to 1.36 atm as the system reached equilbrium. Calculate Kp (in atm-1) for

CO(g) + Cl2(g) ↔ COCl2(g)

I found the final pressure of the gases to get 94.44 which was really off because I think the answer is supposed to be either single digits or a decimal. Please help!

Thanks in advance!

Answer 1

1.40 mol/8.0 L = 0.175M = (CO)
3.20 mol/8.0L = 0.400 M = (Cl2)

............CO + Cl2 ==> COCl2
I........0.175...0.400.....0
C..........-x.....-x.......x
E.......0.175-x..0.400-x....x

Kc = 2.5 = (COCl2)/(CO)(Cl2).
Solve for x.
If you will post your work I will find the error. You know 0.187 can't be right; that's more CO than you started with.

.........CO + Cl2 ==> COCl2
I.......0.74..0.92.....0
C........-p....-p.......p
E.....0.74-p.0.92-p.....p

Ptotal = 1.36 = 0.74-p + 0.92-p + p
Solve for p = 0.30 and substitute into Kp expression. Solve for Kp.

Answer 2

thank you!